Задание 10 Параграф 17 ГДЗ Рабочая тетрадь 2 Мерзляк Полонский 8 класс (Алгебра)

1) (a^2-6)/(a-v6);
2) (vb+7)/(b+14vb+49);
3) (c-2vc+1)/(c-1);
4) (3+v3)/v3;
5) v18/(2-v2);
6) (25+2v5)/(2v3+5v15);
7) (v6a-v12a)/(2-v8);
8) (ava-11v11)/(va-v11);
9) (cvc+1000)/(c-10vc+100).

1. $$\frac{a^2-6}{a-\sqrt6}=\frac{a^2-(\sqrt6)^2}{a-\sqrt6}=\frac{(a-\sqrt6)(a+\sqrt6)}{a-\sqrt6}=a+\sqrt6.$$
2. $$\frac{\sqrt b+7}{b+14\sqrt b+49}=\frac{\sqrt b+7}{(\sqrt b)^2+2\cdot 7\sqrt b+7^2}=\frac{\sqrt b+7}{(\sqrt b+7)^2}=\frac1{\sqrt b+7}.$$
3. $$\frac{c-2\sqrt c+1}{c-1}=\frac{(\sqrt c)^2-2\cdot \sqrt c\cdot 1+1^2}{(\sqrt c)^2-1^2}=\frac{(\sqrt c-1)^2}{(\sqrt c-1)(\sqrt c+1)}=\frac{\sqrt c-1}{\sqrt c+1}.$$
4. $$\frac{3+\sqrt3}{\sqrt3}=\frac{(\sqrt3)^2+\sqrt3}{\sqrt3}=\frac{\sqrt3(\sqrt3+1)}{\sqrt3}=\sqrt3+1.$$
5. $$\frac{\sqrt{18}}{2-\sqrt2}=\frac{3\sqrt2}{2-\sqrt2}\cdot\frac{2+\sqrt2}{2+\sqrt2}=\frac{3\sqrt2(2+\sqrt2)}{4-2}=\frac{3\sqrt2(2+\sqrt2)}{2}.$$
   
   После сокращения:
   $$\frac{\sqrt{18}}{2-\sqrt2}=3+3\sqrt2.$$
6. $$\frac{25+2\sqrt5}{2\sqrt3+5\sqrt{15}}=\frac{5(\sqrt5)^2+2\sqrt5}{2\sqrt3+5\sqrt5\cdot\sqrt3}=\frac{\sqrt5(5\sqrt5+2)}{\sqrt3(2+5\sqrt5)}=\frac{\sqrt5}{\sqrt3}.$$
7. $$\frac{\sqrt{6a}-\sqrt{12a}}{2-\sqrt8}=\frac{\sqrt{6a}-\sqrt{6a\cdot 2}}{2-2\sqrt2}=\frac{\sqrt{6a}(1-\sqrt2)}{2(1-\sqrt2)}=\frac{\sqrt{6a}}{2}.$$
8. $$\frac{a\sqrt a-11\sqrt{11}}{\sqrt a-\sqrt{11}}=\frac{(\sqrt a)^3-(\sqrt{11})^3}{\sqrt a-\sqrt{11}}.$$
   
   По формуле разности кубов:
   $$\frac{(\sqrt a-\sqrt{11})(a+\sqrt{11a}+11)}{\sqrt a-\sqrt{11}}=a+\sqrt{11a}+11.$$
9. $$\frac{c\sqrt c+1000}{c-10\sqrt c+100}=\frac{(\sqrt c)^3+10^3}{c-10\sqrt c+100}.$$
   
   По формуле суммы кубов:
   $$\frac{(\sqrt c+10)(c-10\sqrt c+100)}{c-10\sqrt c+100}=\sqrt c+10.$$

Ответ

1) $$a+\sqrt6$$; 2) $$\frac1{\sqrt b+7}$$; 3) $$\frac{\sqrt c-1}{\sqrt c+1}$$; 4) $$\sqrt3+1$$; 5) $$3+3\sqrt2$$; 6) $$\frac{\sqrt5}{\sqrt3}$$; 7) $$\frac{\sqrt{6a}}{2}$$; 8) $$a+\sqrt{11a}+11$$; 9) $$\sqrt c+10$$.